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GENES SEGREGATION BY BEAN-OTYPEI. Objectives: At the end of the activity, the students should be able to: Simulate the sorting and recombination of genes that occurs during fertilization; Learn to compute and apply Chi-square test on observed phenotype frequency; Derive genotypes based on observed p...

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GENES SEGREGATION BY BEAN-OTYPE

I. Objectives: At the end of the activity, the students should be able to: 1. Simulate the sorting and recombination of genes that occurs during fertilization; 2. Learn to compute and apply Chi-square test on observed phenotype frequency; 3. Derive genotypes based on observed phenotype segregation ratios; and 4. Illustrate symbolically how Mendelian inheritance occurs. II. Discussion: A Mendelian trait is caused by a single gene. Modes of inheritance reveal whether a Mendelian trait is dominant or recessive and whether the gene that contains it is carried on autosome or a sex chromosome. Mendel’s first law, which can predict the probability that a child will inherit a Mendelian trait, applies a new to its child. Mendelian Inheritance of gene applies the Law of Segregation and the Law of Independent Assortment. III. Materials: Colored papers cut into small pieces to represent the gene for a certain trait of garden pea. (60 pieces of the same size and shape; 30 pieces of which is with a different color to the other 30 pieces Colored/Labeled containers (will serve as parent containers) IV. Procedure: 1. Place the 30 pieces of paper cuts (one color) to one container and the other 30 pieces to the other container. Assign which colored paper will represent dominant and recessive genes of a specific trait of garden pea. 2. Label the container as First Parent P1 male and other P1 female. 3. At random, segregate the beans on the Lab table. Assume fertilization occurs and First filial generations (F1) are formed. 4. Categorize the formed First filial generations.

5. Put all the papers in one container. Assume that all first filial generations undergo self-fertilization. 6. Pick-up the papers by two from the container without looking at it and collect the second Filial generation (F2) 7. Categorize the formed Second filial generations into two (2) phenotype, and three (3) genotype classes. 8. Fill up Table 15.1 for the genotypic result, and Table 15.2 for the phenotypic analysis. 9. Compute for the Chi-Square Value and compare it with the Tabular Value. 10.Interpret the results of your Chi-square test. Determine if null hypothesis is accepted or rejected. 11. Diagram the cross showing segregation of genes influencing each phenotype or genotype.

Name _____________________________

Laboratory Instructor _____________

Course/Year/Sec ___________________

Date ___________________________

Activity Sheet _____ GENES SEGREGATION BY BEAN-OTYPE A. Genotypic Analysis 1. The Problem

generations Dominant trait,

Fertilization of a garden pea plant occurs and First filial generations (F1) are formed. Assuming that all first filial generations undergo self-fertilization. Categorize the formed Second filial into 3 genotypes. Note that white color represents brown color represents recessive trait.

2. The Null Hypothesis:

The genotypic ratio of second Filial generation of garden pea is 1 PP : 2 Pp :1 pp (PP = 25%, Pp = 50%, pp = 25%)

Table 1. Distribution and Chi-square test of the Genotypes of the Beans

Genotype Class

Observed Frequency (O)

Expected Frequency (E)

Deviation (d = O – E)

d2 / E

PP

7

7.5

- 0.5

0.033

Pp

16

15

1

0.067

pp

7

7.5

- 0.5

0.033

Total

30

χ2 =0.133

B. Phenotypic Analysis 1. The Problem

generations Dominant trait,

Fertilization of a garden pea plant occurs and First filial generations (F1) are formed. Assuming that all first filial generations undergo self-fertilization. Categorize the formed Second filial into 2 Phenotype. Note that white color represents brown color represents recessive trait.

2. The Null Hypothesis: The phenotypic ratio of second Filial generation of garden pea is 3:1 (White Bean = 75%, Brown Bean= 25%) Table 2. Distribution and Chi-square test of the Phenotypes of the Beans Observed Frequency (O)

Expected Frequency (E)

Deviation (d = O – E)

White Bean

23

22.5

0.5

0.01

Brown Bean

7

7.5

-0.5

0.033

Phenotype Class

Total

30

d2 / E

χ2 =0.043

V. Questions: 1.

Interpret the result of your Chi-square test.

• The null hypothesis is accepted if the value of the computed chi-square is less than the tabulated value. • The null hypothesis is rejected if the value of the computed chi-square is greater than the tabulated value.

Table 1 Computed Value Tabulated Value χ2 =0.133 5.991 Accept null hypothesis

Table 2 Computed Tabulated Value χ2 =0.043 3.841 Accept null hypothesis

2.

Diagram the cross showing segregation of genes influencing each character. Determine the phenotypic ratio (P.R.) and the genotypic ratio (G.R.) of the crosses.

P1 to F1

P2 to F2

P1

PP

pp

gametes

P

p

F1

(ALL)

Pp

Genotypic ratio All Pp Phenotypic ratio All White

Documentation

Pp

gametes

½p

½P

P

p

P

PP

Pp

p

Pp

pp

F2

Genotypic ratio 1 PP : 2Pp :1pp Phenotypic ratio 3 white : 1 brown...